ON THE CHROMATIC NUMBER OF SUBSETS OF THE EUCLIDEAN PLANE
M. AXENOVICH, J. CHOI, M. LASTRINA, T. MCKAY, J. SMITH, AND B. STANTON
Abstract. The chromatic number of a subset of the real plane is the smallest number of colors assigned to the
elements of that set such that no two points at distance 1 receive the same color. It is known that the chromatic
number of the plane is between 4 and 7. In this note, we determine the bounds on the chromatic number for
several classes of subsets of the plane such as extensions of the rational plane, sets in convex position, infinite
strips, and parallel lines.
1. Introduction
Let (X, d) be a graph with vertex set X ⊆ R2 , and u ∼ v if and only if the Euclidean distance || · || between
them is d. When d = 1, this graph is called a unit distance graph. It is well known that 4 ≤ χ(R2 , d) ≤ 7, for
any positive d. In the literature, χ(R2 , 1) is referred to as the chromatic number of the plane, and the problem
of determining its exact value is the famed Hadwiger-Nelson problem. The body of work on this problem is
extensive (see for example expository works [16], [4]).
One important result is the following theorem, which allows the problem to be reduced to finite subsets of
the plane.
Theorem 1.1 (de Bruijn and Erdős [5]). If k is a positive integer and G is a graph such that any finite subgraph
is k-colorable, then χ(G) ≤ k.
In this paper, we examine special subsets of the plane and investigate the chromatic number of the corresponding graph. In Section 2, we bound χ(X, 1), where X belongs to a class of subsets of Q × R. In Section
3, we prove that χ(X, 1) ≤ 4 if X corresponds to the vertex set of a convex |X|-gon. Finally, in Section 4, we
investigate the chromatic number of infinite strips and unions of lines.
2. On the chromatic number of subsets of Q × R
The fact that χ(Q2 , 1) = χ(Z2 , d) = 2 for any natural d was first shown by Woodall [17], and this result has
since been reproven independently by many others. A natural next question is determining χ(Q × R, 1). We
√ know that χ(Q × R, 1) ≥ 3, since the triangle with vertices (0, 0), (1, 0), and 21 , 23 forms a subgraph of this
graph. Is it true that χ(Q × R, 1) = 4? Is it true that χ(Q × R, 1) < χ(R2 , 1)? While we cannot answer these
questions, we provide some partial results.
Date: April 21, 2010.
2010 Mathematics Subject Classification. Primary 05C12, 05C15, 05C62, 51K99.
Supported in part by NSA grant H98230-09-1-0063 and NSF grant DMS-0901008.
1
For the following discussion, let N∗ be the set of square-free natural numbers, i.e., the natural numbers
√
that are not divisible by 4, 9, 16, etc. For distinct N1 , . . . , Nk ∈ N∗ , let Z(N1 , N2 , . . . , Nk ) = {a1 N1 + · · · +
√
ak Nk : a1 , . . . , ak ∈ Z}.
Theorem 2.1. χ(Q × R, 1) = max{χ(Z × Z(N1 , N2 , . . . , Nk ), d) : N1 , . . . , Nk ∈ N∗ , d, k ∈ N}.
Proof. Let H be a finite connected subgraph of (Q×R, 1) with vertex set {v0 = (0, 0), v1 = (p1 /q1 , x1 ), . . . , vn =
(pn /qn , xn )} ⊆ Q × R, where pi , qi ∈ Z and qi 6= 0 for i = 1, . . . , n. Let d be the least common multiple of the qi ,
1 ≤ i ≤ n. Consider the graph H 0 on vertex set dv1 , . . . , dvn , where two vertices are adjacent if and only if the
Euclidean distance between them is d. We see that H is isomorphic to H 0 , and H 0 is a subgraph of (Z × R, d)
that also contains the origin.
We now show that each vertex of H 0 is (x, y), where x ∈ Z and y is an integer combination of square roots of
positive integers. Let 0 = u0 , u1 , . . . , um be a path in H 0 such that u0 , . . . , um−1 have such a representation,
√
√
and consider um . We have that ||um−1 − um || = d. Let um−1 = (b, a1 N1 + · · · + ak Nk ) and um = (b0 , y),
for b0 , b, a1 , . . . , ak ∈ Z, and y ∈ R. Then
p
p
(b − b0 )2 + (a1 N1 + · · · + ak Nk − y)2 = d2 .
p
√
√
So, a1 N1 + · · · + ak Nk − y = ± d2 − (b − b0 )2 . Since d2 − (b − b0 )2 ∈ N, we have that y is also an integer
combination of square roots of positive integers, as desired. To complete the proof, we apply Theorem 1.1. A lot of work has been done on determining χ(Z × Z(N ), d) for N ∈ N∗ and d ∈ N. An interesting question
is whether χ(Z × Z(N1 , N2 , . . . , Nk ), d) ≤ max{χ(Z × Z(N ), d) : N ∈ N∗ }, for any N1 , . . . , Nk ∈ N∗ and d ∈ N.
Theorem 2.2 summarizes the results of Johnson [11], Fischer [9], Moorehouse [14] and others.
Theorem 2.2. Let N and d be positive integers. Then


 2, if N 6≡ 3 (mod 4);



 3, if N 6≡ 2 (mod 3);
χ(Z × Z(N ), d) ≤

4, if N ≡ 3 (mod 8), N 6≡ 47, 59, 83 (mod 84), or





N 6≡ 3, 5, 6 (mod 7).
Here, we prove a more general statement.
Theorem 2.3. If d, k ∈ N and N1 , . . . , Nk ∈ N∗ , then


2, if d 6≡ 0 (mod




 3, if d 6≡ 0 (mod
χ(Z × Z(N1 , . . . , Nk ), d) ≤

4, if d ≡ 2 (mod




 6, if d 6≡ 0 (mod
2) or if Ni 6≡ 3 (mod 4) for 1 ≤ i ≤ k;
3) or if Ni 6≡ 2 (mod 3) for 1 ≤ i ≤ k;
4) or if Ni 6≡ 7 (mod 8) for 1 ≤ i ≤ k;
12) or if Ni 6≡ 11 (mod 12) for 1 ≤ i ≤ k.
Proof of Theorem 2.3. The proof is based on several observations concerning number theoretic properties of
vertices of the graph under consideration. Let V = Z × Z(N1 , . . . , Nk ), where N1 , . . . , Nk ∈ N∗ . We state
several claims and provide their proofs in the Appendix.
2
Claim 1. Let u, v ∈ V , such that ||u − v|| = d, an integer. Then there exist N ∈ {N1 , N2 , . . . , Nk } and
√
a, b ∈ Z such that u − v = (a, b N ).
Claim 2. Let γ, q, a, b, N ∈ Z, d = 2γ q, where γ ≥ 0, q is odd, N ∈ N∗ , and a2 + b2 N = d2 . Then
(1) (γ = 0) =⇒ (a + bN 6≡ 0 (mod 2));
(2) (γ ≥ 1) =⇒ (a ≡ b (mod 2));
(3) (γ ≥ 1 and N 6≡ 3 (mod 4)) =⇒ (a and b are multiples of 2γ );
(4) (γ ≥ 2 and N 6≡ 7 (mod 8)) =⇒ (a and b are multiples of 2γ−1 ).
Claim 3. Let γ, q, a, b, N ∈ Z, d = 3γ q, where γ ≥ 0, q is not a multiple of 3, N ∈ N∗ , and a2 + b2 N = d2 .
Then
(1) (γ = 0) =⇒ (a + bN 6≡ 0 (mod 3));
(2) (γ ≥ 1) =⇒ (a2 ≡ b2 (mod 3));
(3) (γ ≥ 1 and N 6≡ 2 (mod 3)) =⇒ (a and b are multiples of 3γ ).
Now we proceed with the proof of Theorem 2.3. Let G be a connected component of (V, d) and assume without
loss of generality that it contains the origin. In each of the cases addressed by the theorem, we provide an
explicit coloring, c, of G, and check that it is proper, taking u, v ∈ V such that ||u − v|| = d and verifying that
c(u) 6= c(v).
√
By Claim 1, u − v = (a, b N ), where a, b ∈ Z and N ∈ {N1 , . . . , Nk }. Moreover, d2 = a2 + b2 N .
For each vertex, we introduce the notation:
v = (av ,
k
X
bvi
p
Ni ).
i=1
(1) (a) Suppose d is odd. Let c(v) = av +
Pk
i=1 bvi Ni
(mod 2). Then d ≡ d2 (mod 2) ≡ a2 + b2 N
(mod 2) ≡ a + bN ≡ |c(v) − c(u)| (mod 2). Since d is odd, a + bN is odd, thus c(u) 6= c(v).
(b) Suppose Ni 6≡ 3 (mod 4) for i = 1, . . . , k and d = 2γ q, where γ ≥ 1 and q is odd. Using Claim 2 (3)
and the inclusion of the origin in the vertices of G, we see that the coordinates of each vertex in G
are divisible by 2γ , so G is isomorphic to a subgraph of (Z × Z(N1 , . . . , Nk ), q), which is bipartite
by the previous case.
(2) (a) Suppose d is not a multiple of 3. Let c(v) = av +
Pk
i=1 bvi Ni
(mod 3). From Claim 3 (1), a+bN 6≡ 0
(mod 3). Thus c(u) 6= c(v).
(b) Suppose Ni 6≡ 2 (mod 3) for i = 1, . . . , k and d = 3γ q, where γ ≥ 1 and q is not a multiple of
3. Since (0, 0) ∈ V (G), Claim 3 (3) implies that the coordinates of each vertex of G are divisible
3
by 3γ , so G is isomorphic to a subgraph of (Z × Z(N1 , . . . , Nk ), q), which is 3-colorable from the
previous case.
(3) (a) Suppose d = 2q for some odd q. From Claim 2 (2), a and b have the same parity. Let E(G) =
E1 ∪ E2 , where E1 = {{u, v} ∈ E(G) : au−v ≡ 0 (mod 2)}, E2 = E − E1 . Let Gi be the subgraph
of G such that E(Gi ) = Ei , i = 1, 2. By considering the directed differences between the rational
coordinates of the vertices in G2 , we conclude that there are no odd cycles in this subgraph, and
so χ(G2 ) = 2. Since au−v ≡ 0 (mod 2) for each edge {u, v} in G1 , by translation and Claim
2 (2) each connected component of G1 is isomorphic to a subgraph of (Z × Z(N1 , . . . , Nk ), q), so
χ(G1 ) = 2 by Case 1. Let c1 and c2 be the proper 2-colorings of G1 and G2 , respectively. Then
c(v) = (c1 (v), c2 (v)), v ∈ V (G), is a proper 4-coloring of G, so χ(G) ≤ 4.
(b) Suppose Ni 6≡ 7 (mod 8) for i = 1, . . . , k and d = 2γ q, where γ ≥ 2 and q is odd. Claim 2 (4)
implies that the coordinates of vertices of G are divisible by 2γ−1 , so G is isomorphic to a subgraph
of (Z × Z(N1 , . . . , Nk ), 2q), which is 4-colorable from the previous case.
(4) First, suppose that d 6≡ 0 (mod 6). Then d 6≡ 0 (mod 2) or d 6≡ 0 (mod 3). In either case, we have
a 2- or 3-coloring from above. Now, if d ≡ 6 (mod 12), then d ≡ 2 (mod 4), in which case we have a
4-coloring. Thus, we may assume that d ≡ 0 (mod 12), and that Ni 6≡ 11 (mod 12) for i = 1, . . . , k.
Define
A = {A1 , . . . , Am } = {Ni : Ni 6≡ 3
(mod 4)} and B = {B1 , . . . , Bm0 } = {N1 , . . . , Nk } \ A.
Note that Bi 6≡ 2 (mod 3) since Ni 6≡ 11 (mod 12), and so Bi ≡ 3 or 7 (mod 12), i = 1, . . . , m0 . For
√
√
each vertex v, we may uniquely write v = v1 + v2 , where v1 = (a0 , a1 A1 + · · · + am Am ) and
√
√
v2 = (0, b1 B1 + · · · + bm0 Bm0 ). Note that from the previous parts, there is a proper 2-coloring
c1 of (Z × Z(A1 , . . . , Am ), d) and a proper 3-coloring c2 of (Z × Z(B1 , . . . , BN ), d). We now define
c(v) = (c1 (v1 ), c2 (v2 )). Clearly, c is a 6-coloring. Now, suppose that u = u1 + u2 is adjacent to v.
√
Then u − v = (a, b Ni ) for some i. If Ni = Aj for some j, then c1 (u1 ) 6= c1 (v1 ). Otherwise, Ni = Bj
for some j and so c2 (u2 ) 6= c2 (v2 ). In either case, we have c(u) 6= c(v) and so c is a proper 6-coloring.
3. On the chromatic number of (X, 1), where X is a set of points forming a convex |X|-gon
Let V be a set of n points in convex position in the plane. Let f (V ) be the number of pairs of points in
V at distance 1. Let f (n) = max{f (V ) : |V | = n, V ⊆ R2 , V is in convex position}. It was conjectured that
f (n) ≤ 2n − 2, by Erdős and Moser [8], Erdős and Fishburn [7]. To date, the best upper bound on f (n) is
Cn log n for a constant C, and is due to Füredi [10] and Braß and Pach [3]. The best lower bound, f (n) ≥ 2n−7,
is due to Edelsbrunner and Hajnal [6]. As we investigate this problem, we shall interpret most of the results in
terms of a graph G(V ) with vertex set V and an edge set consisting of the pairs of vertices at distance 1. Thus
4
Xi
Xi’
X’
Y’
Yj
Yj’
S
Y
X
B
Figure 1. Two points X, Y at a maximum distance and their neighbors.
f (V ) = |E(G(V ))|. Here, we investigate this problem. Let Br (A) be a ball of radius r with center A, and let
Sr (A) be a circle of radius r with center A.
Lemma 3.1. Let V be a subset of the plane in convex position with diameter greater than
√
2, then there is a
vertex of degree at most 3 in G(V ).
Proof. Let d be the diameter of V , d >
√
2. Consider two points X, Y at distance d. Assume, without loss of gen-
erality, that the line through X and Y is horizontal. Assume that the degree of each vertex in G(V ) is at least 4.
Idea of the proof:
Consider 4 neighbors of X and 4 neighbors of Y . We shall first observe that at most two of the neighbors of X
could be in the “middle” arc of the unit circle centered at X, otherwise V is not in a convex position. Next,
we observe that if a neighbor of X is in the “upper” arc, and a neighbor of Y is in the “lower” arc of the
corresponding circles, then the distance between these neighbors is greater than d, contradicting the diameter
condition. Thus, without loss of generality there must be two neighbors of X and two neighbors of Y in the
“upper” part of the corresponding circles. See Figure 1. These four points, together with X and Y are not in
convex position, resulting in a final contradiction.
5
For a formal argument, we introduce some notation. Let, the sets of neighbors, N (X) = {X1 , . . . , Xp } ⊆
S1 (X) and N (Y ) = {Y1 , . . . , Yq } ⊆ S1 (Y ) with p, q ≥ 4. Since diam(V ) = d, we have that V ⊆ B =
Bd (X) ∩ Bd (Y ). Let us denote the half-plane above line XY as Pup , and the half-plane below XY as Pdown .
Let X 0 , X 00 , Y 0 , Y 00 be “tangent” points, Y 0 ∈ S1 (Y ) ∩ Pup , X 0 ∈ S1 (X) ∩ Pup , Y 00 ∈ S1 (Y ) ∩ Pdown , X 00 ∈
S1 (X) ∩ Pdown such that XY 0 and XY 00 are tangent to S1 (Y ), Y X 0 and Y X 00 are tangent to S1 (X). These
tangent points X 0 and X 00 split S1 (X)∩B into the three arcs mentioned above, and so do Y 0 , Y 00 with S1 (Y )∩B.
Let S1 (X) ∩ B = Su (X) ∪ Sm (X) ∪ Sl (X), let S1 (Y ) ∩ B = Su (Y ) ∪ Sm (Y ) ∪ Sl (Y ), where u, m, l stand
for upper, middle, lower parts, and Sm (X) is the X 0 − X 00 arc of S1 (X), Su (X) is the arc of S1 (X) ∩ B above
Sm (X), and Sl (X) is the arc of S1 (X) ∩ B below Sm (X). We define Su (Y ), Sm (Y ), Sl (Y ) similarly. Note that
XY 0 Y X 00 and XX 0 Y Y 00 are rectangles with diagonals of length d and one of the side lengths 1.
Claim: The distance between Sl (X) and Su (Y ) is greater than d.
To prove this claim, consider two lines `x , `y - one passing through X 00 , another through Y 0 such that they
are perpendicular to X 00 Y 0 . The distance between these lines is d. We see that for d > 1, the slopes of `x and
`y are negative. Thus Sl (X) and Su (Y ) are outside of the plane region between `x and `y , and so the distance
between Sl (X) and Su (Y ) is greater than the distance between `x and `y , namely d.
Observe that there is at most one point of V in each of Sm (Y ) ∩ Pup , Sm (Y ) ∩ Pdown , Sm (X) ∩ Pup , and
Sm (X) ∩ Pdown because of convexity. Since |N (X)|, |N (Y )| ≥ 4, and because of the claim, there are, without
loss of generality, Xi , Xi0 ∈ Su (X), and Yj , Yj 0 ∈ Su (Y ). Note that the slope of Xi Xi0 line is negative and the
√
slope of Yj Yj 0 is positive. This contradicts the convexity of V since, for d > 2, Su (X) is completely to the left
of Su (Y ).
Corollary 3.2. Let V be a subset of a plane in convex position, then χ(G(V )) ≤ 4.
Proof. If V has diameter at most
√
2, it could be embedded into a square with side length
√
2. This square
could be four-colored by splitting in into four squares s1 , s2 , s3 , s4 of equal sizes and assigning color i to each
point of square si .
4. On the chromatic number of (X, 1), where X ⊂ R2 for X an infinite strip and for X a union
of lines
Pritikin [15] produced a 7-coloring of the plane such that one color class, say of color 0, is small. Using this
coloring, it was shown that any 6197 points can be simultaneously translated so that none of the points are
assigned color 0. In fact, a stronger statement was proven.
A good 7-coloring of the plane is a coloring using colors 0, 1, . . . , 6, such that if there are two vertices at
distance 1 having the same color, this color is 0.
6
Figure 2. A tiling of the plane with tiles of 7 different colors, the tiles in each column use 3
different colors, every other column has the same coloring (shifted)
Theorem 4.1 (Pritikin [15]). Let c be a good 7-coloring of the plane which is periodic, i.e., there are α, β, γ > 0
such that for all (x, y) ∈ R2 , c(x, y) = c(x, y + β) = c(x + α, y + γ). Let S = {(x, y) : 0 ≤ x < α, 0 ≤ y <
β, c(x, y) = 0} be a Riemann integrable set of area A > 0. If |X| < αβ/A, then χ(X, 1) ≤ 6.
The main idea of Pritikin’s construction is to consider a pentagonal tiling of the plane (of type 5, according
to Wolfram Alpha classification of 14 known pentagonal tilings), color it periodically with colors 1, . . . , 6, and
then recolor small diamond-shaped regions with the color 0. See Figure 2 for an illustration.
In [1] and [13], the chromatic numbers of regions of the plane, such as circular or rectangular regions, are
investigated. Several interesting precise results are proven in those papers with regards to 3-colorable regions.
In particular, Bauslaugh [1] proves that an infinite strip in the plane is 3-colorable if and only if it has width
√
at most 3/2. Finding such a sharp result for 4-colorable strips seems to be difficult, and investigating 5 and
6-colorable strips is probably as difficult as the original question. However, we can provide the lower bounds on
the width of an i-colorable strip in the plane, for i = 4, 5, 6 using constructions. Let ω(i) be the largest width
of an i-colorable strip in the plane.
Theorem 4.2.
Proof. The fact that ω(3) ≥
√


3/2 ≥ 0.866,




√

2 2/3 ≥ 0.94,
ω(i) ≥ √


15/4 ≥ 0.968,




√

 15/2 + √3 ≥ 3.668,
√
i = 3,
i = 4,
i = 5,
i = 6.
3/2 is a theorem proved by Bauslaugh [1]. To see that ω = ω(4) ≥
p
8/9, as
mentioned in [1], construct the colorings of an infinite strip of width ω built of monochromatic ω ×1/3 rectangles
7
p
of colors 1, 2, 3, 4, 1, 2, 3, 4, . . ., see Figure 3. To see that ω = ω(5) ≥ 15/16, build a similar construction with
√
√
monochromatic ω × 1/4 rectangles. To observe that ω(6) ≥ 15/2 + 3, we consider a coloring, similar to
one used by Pritikin, see Figure 4. The symmetric tiles have width 1/2, with the rounded boundary being a
circular arc of radius 1 centered at the opposite corner of the tile. The tiles used in the top and bottom rows
are determined by the distance 1 between the point of intersection of circular arcs and the opposite corner of a
tile.
Figure 3. A coloring of an infinite strip using 4 colors
Figure 4. A coloring of an infinite strip using 6 colors
Theorem 4.3. Let X be the union of two parallel lines on the plane. If the distance ` between two lines satisfies
one of
8
a) ` ≥ 1,
b) ` < 1 and
c) ` < 1 and
√
√
1 − `2 is irrational, or
1 − `2 is rational with an odd denominator in its lowest term,
then χ(X, 1) = 2. Otherwise, χ(X, 1) = 3.
Proof. Assume that X is the union of vertical lines L1 and L2 at distance `. If ` > 1, then we can color
L1 and L2 in two colors independently. If ` = 1, then, for each vertex in L1 , there is exactly one vertex in
L2 adjacent to it, namely, a vertex with the same y-coordinate. Let c be a proper coloring of (L1 , 1) with
colors 0 and 1. For x ∈ L2 , let c(x) = 1 − c(x0 ), where x0 ∈ L1 and x0 has the same y-coordinate as x. This
gives a proper 2-coloring of (X, 1) when ` = 1. So, assume for the rest of the proof that ` < 1. Consider a
finite subset X 0 of X and a point in X 0 with the largest y-coordinate. It is adjacent to at most 2 other vertices
in (X 0 , 1). By induction on the size of X 0 , it follows that χ(X 0 , 1) ≤ 3. Theorem 1.1 then gives that χ(X, 1) ≤ 3.
We shall now prove that χ(X, 1) = 2 when ` satisfies conditions b) or c) of the theorem. Assume that
(0, 0) ∈ L1 . Let L be the connected component of (X, 1) containing (0, 0). Then the y-coordinate, uy , of any
√
point u in V (L) can be expressed in the form uy = a + b 1 − `2 , where a, b ∈ Z, and b is even if u ∈ L1 , b is
odd if u ∈ L2 . This fact can be easily verified by induction on |V (L)|. We shall refer to such a representation
uy of u as a good representation.
Case 1. Either
√
1 − `2 is irrational or
√
1 − `2 = p/q, p, q ∈ Z, q-odd.
√
√
Claim. For any two good representations of a point in L, a + b 1 − `2 = a0 + b0 1 − `2 , we have a ≡ a0
(mod 2) and b ≡ b0 (mod 2).
√
Since the representations are good, we have immediately that b ≡ b0 (mod 2). Let 1 − `2 be irrational. Since
√
a − a0 + (b − b0 ) 1 − `2 = 0, the linear independence of irrational numbers over rational numbers gives that
√
√
b = b0 and thus a = a0 . Let 1 − `2 = pq , where p, q ∈ Z and q is odd. So, a − a0 + (b − b0 ) 1 − `2 = 0, and
q(a − a0 ) + (b − b0 )p = 0. Since q is odd and b − b0 is even, a − a0 must be even. This verifies the claim.
√
For a vertex u with good representation a + b 1 − `2 , let the color of u, c(u) = a + b (mod 2). From the
Claim, the coloring is well-defined. Since any points on the same line Li have the same parity for a coefficient in
√
front of 1 − `2 , it gives a proper coloring of Li , i = 1, 2. Assume that u ∈ L1 , v ∈ L2 are two adjacent vertices
√
√
of the same color. Let uy and vy be good representations of u and v: uy = a + b 1 − `2 , vy = a0 + b0 1 − `2 .
Note that since b and b0 have different parities and c(u) = c(v), a and a0 must have different parities. Since
√
√
||u − v||2 = 1 = `2 + ((a − a0 ) + (b − b0 ) 1 − `2 )2 , we have that (a − a0 ) + (b − b0 ± 1) 1 − `2 = 0.
√
√
If 1 − `2 is irrational and a 6= a0 , we have a contradiction. If 1 − `2 = pq , p, q ∈ Z and q is odd, then
(±1 − b + b0 ) pq = (a − a0 ), so (±1 − b + b0 )p = q(a − a0 ). Here, the left-hand side is even and the right-hand side
is odd, a contradiction.
9
Case 2.
√
1 − `2 =
p
2q ,
p, q ∈ Z, gcd(p, 2q) = 1.
√
√
√
Observe that L contains an odd cycle with vertices (0, 0), (`, 1 − `2 ), (0, 2 1 − `2 ), . . ., (`, (2q − 1) 1 − `2 ),
√
(0, 2q 1 − `2 ) = (0, p), (0, p − 1), . . ., (0, 1), (0, 0), so χ(X, 1) ≥ 3 in this case.
Acknowledgements The authors would like to thank the Department of Mathematics at Iowa State University for supporting this graduate research project and Felix Lazebnik for bringing the manuscript of Moorehouse
to their attention.
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5. Appendix - Proofs of Claims from Section 2
√
Pk
Proof of Claim 1. We have that w = u − v is a vector of length d, d2 = a2w + ( i=1 bwi Ni )2 . Thus
√
Pk
2
2
i=1 bwi Ni is the square root of an integer, d − aw . We know that any set consisting of distinct square free
integers is linearly independent over the field Q (this is from a simple fact of quadratic field extensions, or see,
for example, [2], [12]). Thus all but at most one expression bwi is zero.
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Proof of Claim 2.
(1) We have that a + bN ≡ a2 + b2 N = d2 ≡ d ≡ 1 (mod 2).
(2) We have that a2 + b2 N ≡ 0 (mod 4). Since N is square free, a is even if and only if b is even.
(3) Suppose a and b are odd. Then, from a2 + b2 N = d2 , we have 1 + N ≡ 0 (mod 4), which is impossible if
N 6≡ 3 (mod 4). Hence from (2), a and b are even. By induction on γ ≥ 1, we are done.
(4) Similar to (3), but consider modulo 8 and induction on γ ≥ 2.
Proof of Claim 3
(1) Suppose d 6≡ 0 (mod 3). Hence d2 ≡ 1 (mod 3). So a2 + b2 N ≡ 1 (mod 3). If N ≡ 0 (mod 3), then
a2 ≡ 1 (mod 3), i.e. a 6≡ 0 (mod 3). Thus a + bN 6≡ 0 (mod 3). If N ≡ 1 (mod 3), then a2 + b2 ≡ 1 (mod 3),
i.e. (a2 ≡ 1 (mod 3) and b2 ≡ 0 (mod 3)) or (a2 ≡ 0 (mod 3) and b2 ≡ 1 (mod 3)). In either case, we have
a + bN 6≡ 0 (mod 3). If N ≡ 2 (mod 3), then a2 + 2b2 ≡ 1 (mod 3), i.e. a2 ≡ 1 (mod 3) and b2 ≡ 0 (mod 3).
So a 6≡ 0 (mod 3) and b ≡ 0 (mod 3). Therefore we have a + bN 6≡ 0 (mod 3).
(2) Suppose a2 6≡ b2 (mod 3). So one of a and b is a multiple of 3. Since N is square free, the other should be
a multiple of 3, i.e. a2 ≡ b2 ≡ 0 (mod 3), a contradiction.
(3) Suppose d ≡ 0 (mod 3) and N 6≡ 2 (mod 3). Hence a2 + b2 N ≡ 0 (mod 3). If a2 ≡ b2 ≡ 1 (mod 3), then
we have 1 + N ≡ 0 (mod 3), which is impossible since N 6≡ 2 (mod 3). From (2), a2 ≡ b2 ≡ 0 (mod 3), i.e.
a ≡ b ≡ 0 (mod 3). Then by induction on γ ≥ 1, we are done.
Department of Mathematics, Iowa State University, Ames, IA 50011
E-mail address: [email protected], [email protected], [email protected], [email protected], [email protected],
[email protected]
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